Question: A certain circle can be represented by the following equation. $x^2+y^2-8x+8y-49=0$ What is the center of this circle ? $($
Explanation: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2-8x+8y-49&=0\\\\ x^2+y^2-8x+8y&=49\\\\ (x^2-8x)+(y^2+8y)&=49 \text{(rearrange terms)}\\\\ (x^2-8x{+16})+(y^2+8y{+16})&=49{+16}{+16}\end{aligned}$ Notice that we must add ${16}$ and ${16}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 16 and 16?] Writing the equation in standard form $\begin{aligned}(x^2-8x{+16})+(y^2+8y{+16})&=49{+16}{+16}\\\\ (x-{4})^2+(y+4)^2&=81\\\\ (x-4)^2+(y-(-4))^2&=9^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(4,-4)$ and has a radius of $9$ units. Summary The circle is centered at $(4,-4)$. The circle has a radius of $9$ units.